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STRUCTURE OF BERYLLIUM ISOTOPES
By Prof. Lefteris Kaliambos (Natural Philosopher in New Energy) (August 2014) Historically the discovery of the assumed uncharged neutron (1932) along with the invalid relativity (EXPERIMENTS REJECT RELATIVITY) led to the abandonment of the well-established electromagnetic laws, in favour of various contradicting nuclear theories which cannot lead to the nuclear structure. Under this physics crisis in 2003 I published my paper “Nuclear structure is governed by the fundamental laws of electromagnetism ” which led to my discovery of the new structure of protons and neutrons given by proton = + 5d + 4u = 288 quarks = mass of 1836.15 electrons neutron = + 4u + 8d = 288 quarks = mass of 1838,68 electrons The paper was also presented at a nuclear conference held at NCSR "Demokritos" (2002). Here one can see the 9 charged quarks in proton and the 12 ones in neutron able to give the charge distributions in nucleons for revealing the strong electromagnetic force for the nuclear binding in the correct nuclear structure by applying the laws of electromagnetism. You can see my papers of nuclear structure in my FUNDAMENTAL PHYSICS CONCEPTS . Note that according to my discovery of the LAW OF ENERGY AND MASS the mass defect in the nuclear structure is due to the photon mass of the emitting dipolic photon presented at the international conference "Frontiers of fundamental physics" (1993) organised by the natural philosophers M. Barone and F. Selleri , who gave me an award including a disc of the atomic philosopher Democritus. Nevertheless today many physicist continue to apply not the well-established laws but the various fallacious nuclear structure models which lead to complications. Beryllium (Be) has 12 known isotopes, but only one of these isotopes (Be-9) is stable and a primordial nuclide. As such, beryllium is considered a monoisotopic element. It is also a mononuclidic element, because its other isotopes have such short half-lives that none are primordial and their abundance is very low. Beryllium is unique as being the only monoisotopic element with both an even number of protons and an odd number of neutrons. There are 25 other monoisotopic elements but all have odd atomic numbers, and even numbers of neutrons. Of the 11 radioisotopes of beryllium, the most stable are Be-10 with a half-life of 1.39 million years and Be-7 with a half-life of 53.22 days. All other radioisotopes have half-lives under 13.85 seconds, most under 20 milliseconds. The least stable isotope is Be-6, with a half-life measured as 5.03 leptoseconds. The natural light-element ratio of equal protons and neutron numbers is prevented in beryllium by the extreme instability of Be-8 toward alpha decay, which is favored due to the extremely tight binding of He-4 nuclei. The half-life for the decay of Be-8 is only 6.7(17)×10−17 seconds. For understanding better the structure of the unstable Be-8 you can see the following second figure. Beryllium is prevented from having a stable isotope with 4 protons and 6 neutrons by the very large mismatch in proton/neutron ratio for such a light element. Nevertheless, this isotope, Be-10, has a half-life of 1.39 million years, which indicates unusual stability for a light isotope with such a large neutron/proton imbalance. Still other possible beryllium isotopes have even more severe mismatches in neutron and proton number, and thus are even less stable. Most Be-9 in the universe is thought to be formed by cosmic ray nucleosynthesis from cosmic ray spallation in the period between the Big Bang and the formation of the solar system. The isotopes Be-7, with a half-life of 53 days, and Be-10 are both cosmogenic nuclides because they are made on a recent timescale in the solar system by spallation, like C-14. These two radioisotopes of beryllium in the atmosphere track the sun spot cycle and solar activity, since this affects the magnetic field that shields the Earth from cosmic rays. The rate at which the short-lived Be-7 is transferred from the air to the ground is controlled in part by the weather. Be -7 decay in the sun is one of the sources of solar neutrinos, and the first type ever detected using the Homestake experiment. Presence of Be-7 in sediments is often used to establish that they are fresh, i.e. less than about 3–4 months in age, or about two half-lives of Be-7. WHY Be-9 WITH S =-3/2 IS A STABLE NUCLIDE, WHILE THE PARALLELEPIPED OF Be-8 WITH S =0 DECAYS INTO TWO ALPHA PARTICLES After a careful analysis of the structure of atomic nuclei I discovered that the beta decay is due to the fact that in unstable nuclei there exist single pn bonds of weak horizontal bonds leading to the beta decay. For example in my papers STRUCTURE AND BINDING OF H3 AND He3 using the diagram of the structure of the H3 one sees that it is unstable because the two neutrons make single np bonds, while the He3 is stable because the one neutron between the two protons makes two np bonds per neutron. Moreover the pp repulsions of long range lead to the instability when we have a small number of pn bonds per neutron or per proton. In the case of Be-9 using the diagram of Be-9 we see that all neutrons at the corners of the rectangle make two bonds per neutron and the central n3 makes four pn bonds per neutron which contribute significantly to the increase of the binding energies of bonds which overcome the pp and nn repulsions. ' DIAGRAM OF STABLE Be-9 WITH S=-3/2' ' n4 (-1/2)..p4( -1/2)..n5(-1/2) ' ' p2(+1/2)..n3 (+1/2).p3(+1/2) ' ' ''' n1( -1/2)..p1(-1/2)..n2( -1/2) ' It is of interest to note that the antiparallel spin of the pp and nn systems along the radial direction gives magnetic attractions. Under this condition the pp repulsions at the vertical rectangles as in the case of helium-4 are not strong, because we observe electric repulsions and magnetic attractions. .However the simplest parallelepiped of Be-8 decays because the pp repulsions of the two squares overcome the pn bonds. In the following diagram of Be-8 we see that the protons of the same square have parallel spin. Thus the magnetic repulsions of the pp systems are very strong and lead to the decay. '' ''DIAGRAM OF THE UNSTABLE Be-8 WITH S =0''' Here the p1n1n2p2 make the first horizontal square with positive spins at the first horizontal plane (+HP1) , while the n3p3p4n4 make the second horizontal square at the second horizontal plane of negative spins (-HP2). All these nucleons form the first unstable parallelepiped of the structure of atomic nuclei, because the parallel spin at the same square gives strong magnetic repulsions. ' p4(-1/2).......n4(-1/2)' ' n3(-1/2)......p3(-1/2) -HP2' ' n2 (+1/2).....p2(+1/2)' p1(+1/2)......n1(+1/2) +HP1 ''' '''STRUCTURE OF Be-5, Be-6 and Be-7 Starting with the diagram of Be-9 we see that in the absence of the 4 nucleons like the n1,n2 n4 and n5 we get the structure of Be-5 with S = +1/2 as shown in the following diagram. Here we have 3 nucleons of positive spins and 2 nucleons of negative spins giving S = +1/2 . ' DIAGRAM OF Be-5 WITH S = +1/2' ' p4( -1/2) ' ' p2(+1/2)..n3 (+1/2).p3(+1/2) ' ' ''' p1(-1/2) ' Similarly in the absence of 3 nucleons like the n1, n2 and n5 we get the structure of Be-6 with S=0 as shown in the following diagram. Here you see 6 nucleons of opposite spins. ''' DIAGRAM OF Be-6 WITH S =0 n4 (-1/2)..p4( -1/2) ' ' p2(+1/2)..n3 (+1/2).p3(+1/2) ' ' '' ''p1(-1/2) However in the case of Be-7 with S = -3/2 in the absence of 2 neutrons like n2 and n5 of Be-9 we observe a structure with S =-3/2 since the p3(+1/2) of Be-9 changes the spin and goes from n3 to n1 as p3(-1/2) to make a bond with n1 as shown in the following diagram of Be-7 . Here we observe that there are 5 nucleons of negative spins and 2 nucleons of positive spins giving the total spin S=-3/2 . ' DIAGRAM OF Be-7 WITH S =-3/2' n4 (-1/2)..p4( -1/2) ' ' p2(+1/2)..n3 (+1/2) ' ''' ''p3(-1/2) n1( -1/2)..p1(-1/2) TRUCTURE OF Be-10, Be-11, Be-12, Be-13, Be-14, Be-15, Be-16 ''' Adding two extra neutrons of opposite spins like n5(+1/2 and n6(-1/2 in the structure of Be-8 we get the structure of Be-10 with S=0. In the following diagram of Be-10 we see that they ' make single bonds with p1(+1/2) and p3(-1/2) respectively which lead to the beta decay.' ' ' DIAGRAM OF Be-10 WITH S =0''' ' p4(-1/2).......n4(-1/2)' ' n3(-1/2)......p3(-1/2) .....n6(-1/2) -HP2' ' n2 (+1/2).....p2(+1/2)' n5(+1/2)......p1(+1/2)......n1(+1/2) +HP1 ' Then adding the n7(+1/2) at the p2(+1/2) we get the structure of Be-11 with S=+1/2. Whereas the addition of the n8(-1/2) at the p4(-1/2) gives the structure of Be-12 with S=0 as shown in the following diagram of Be-12 ' DIAGRAM OF B-12 WITH S=0 ''' '''n8(-1/2) .............p4(-1/2).......n4(-1/2) ' n3(-1/2)......p3(-1/2) .....n6(-1/2 -HP2' ' n2 (+1/2).....p2(+1/2)......n7(+1/2)' 'n5(+1/2)......p1(+1/2)......n1(+1/2) +HP1 ' Then following the same way we get the structures of B-13 with S=+1/2, of B-14 with S=0 , of B-15 with S=+1/2 and of B-16 with S=0. In these cases the extra neutrons make single bonds with protons existing in front of the protons and behind them. So they are not shown in a diagram. Category:Fundamental physics concepts